Library / Saudi Arabian Mathematical Competitions /112 of 395

Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^{\circ}$ and let $E$ is a point lying on side $CD$ . Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB$ , $AE$ and $BE$ at points $P$ , $F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$ .

Solution

Let $KF$ meets $AB$ at $S$ . We have known that $EP$ , $AK$ , $BF$ are concurrent at Gergonne's point, then $(SP, AB) = -1$ . Let $Q$ be the projection of $P$ on $KF$ then $Q(SP, AB) = -1$ , but $QS \perp QP$ so $QP$ is the angle bisector of $\angle AQB$ . From that, we get $\angle BQM = \angle AQN$ .

In the other hand, easy to realize that $PQMB$ and $PQNA$ are cyclic, so
$\angle BPM = \angle BQM = \angle AQN = \angle APN.$
Combining with the truth that $AB$ is the tangent of $(I)$ , we get
$\angle PTH = \angle BPM = \angle APN = \angle PHT,$
or $PH = PT$ . $\square$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.